Solved – Continuity (from above) theorem

Since you've now offered an explanation of your attempted working (in the comments) I'm going to provide a solution. This property of the probability measure is often referred to as "continuity from above", and it follows as a consequence of countable additivity. The property is usually established via the corresponding property of "continuity from below", but here I will fold that result in to give a proof that only uses the properties of sets and the axioms of probability.

Since $mathcal{A}_1 supseteq mathcal{A}_2 supseteq mathcal{A}_3 supseteq cdots$ we have $overline{mathcal{A}}_1 subseteq overline{mathcal{A}}_2 subseteq overline{mathcal{A}}_3 subseteq …$, which means we can form disjoint sets of the form $overline{mathcal{A}}_k – overline{mathcal{A}}_{k-1}$ for $k geqslant 2$. Using DeMorgan's law and the countable additivity property, we have:

$$begin{equation} begin{aligned} mathbb{P}(mathcal{A}) = mathbb{P} Bigg( bigcap_{k=1}^infty mathcal{A}_{k} Bigg) &= 1 – mathbb{P} Bigg( overline{bigcap_{k=1}^infty mathcal{A}_{k}} Bigg) \[6pt] &= 1 – mathbb{P} Bigg( bigcup_{k=1}^infty overline{mathcal{A}}_k Bigg) \[6pt] &= 1 – mathbb{P} Bigg( overline{mathcal{A}}_1 cup bigcup_{k=2}^infty (overline{mathcal{A}}_k – overline{mathcal{A}}_{k-1}) Bigg) \[6pt] &= 1 – mathbb{P} ( overline{mathcal{A}}_1) + sum_{k=2}^infty mathbb{P} (overline{mathcal{A}}_k – overline{mathcal{A}}_{k-1}) \[6pt] &= 1 – mathbb{P} ( overline{mathcal{A}}_1) + sum_{k=2}^infty [ mathbb{P} (overline{mathcal{A}}_k) – mathbb{P}(overline{mathcal{A}}_{k-1}) ] \[6pt] &= 1 – mathbb{P} ( overline{mathcal{A}}_1) + lim_{n rightarrow infty} sum_{k=2}^n [ mathbb{P} (overline{mathcal{A}}_k) – mathbb{P}(overline{mathcal{A}}_{k-1}) ] \[6pt] &= lim_{n rightarrow infty} Bigg( 1 – mathbb{P} ( overline{mathcal{A}}_1) + sum_{k=2}^n [ mathbb{P} (overline{mathcal{A}}_k) – mathbb{P}(overline{mathcal{A}}_{k-1}) ] Bigg) \[6pt] &= lim_{n rightarrow infty} Big( 1 – mathbb{P} ( overline{mathcal{A}}_n) Big) \[6pt] &= lim_{n rightarrow infty} mathbb{P} ( mathcal{A}_n). \[6pt] end{aligned} end{equation}$$

(The penultimate step in this equation involves recognition of cancelling terms in the sum.)